Question

Для чисел z1=2+3i и z2=1-2i найдите действительное числа а и б, для которых верно z1/z2=az1+bz2

Answer 1

$z_1=2+3i;z_2=1-2i\\ \frac{z_1}{z_2}= \frac{2+3i}{1-2i} = \frac{(2+3i)(1+2i)}{1-4i^2} = \frac{2+4i+3i+6}{1+4} = \frac{8+7i}{5} =az_1+bz_2\\ 8+7i=5a(2+3i)+5b(1-2i)\\ 8+7i=(10a+5b)+(15a-10b)i\\ \left \{ {{10a+5b=8} \atop {15a-10b=7}} \right. \iff \left \{ {{20a+10b=14} \atop {15a-10b=7}} \right. \iff 35a=21\\ \iff a= \frac{21}{35} = \frac{3}{5} =0,6\\ 10*0,6+5b=8 \iff 5b=8-6=2 \iff b= \frac{2}{5} =0,4\\ \underline{ \left \{ {{a=0,6} \atop {b=0,4}} \right. }$